Find the area of the region enclosed by the inner

Find the interval in which the inner loop exists.
For this we take ${\displaystyle r=0}$
${\displaystyle \Rightarrow 2+4\sin \theta =0}$
${\displaystyle \Rightarrow \sin \theta =-\frac{12}{}}$
${\displaystyle \Rightarrow \theta =\frac{7\pi }{6},\frac{11\pi }{6}}$
So inner loop lies between the lines ${\displaystyle \theta =\frac{7\pi }{6}}$ and ${\displaystyle \theta =\frac{11\pi }{6}}$
Since curve is symmetric about the vertical polar axis. So we canf ind the area of the loop by making double the area between ${\displaystyle \theta =\frac{7\pi }{6}}$ and ${\displaystyle \theta =\frac{3\pi }{2}}$
Then Area of the loop is ${\displaystyle A=2{\int }_{7\frac{\pi }{6}}^{3\frac{\pi }{2}}{\frac{12}{r}}^{2}d\theta }$
Therefore
${\displaystyle A={\int }_{7\frac{\pi }{6}}^{3\frac{\pi }{2}}{(2+4\sin \theta )}^{2}d\theta }$
${\displaystyle A=4{\int }_{7\frac{\pi }{6}}^{3\frac{\pi }{2}}{(1+2\sin \theta )}^{2}d\theta }$
${\displaystyle =4{\int }_{7\frac{\pi }{6}}^{3\frac{\pi }{2}}(1+2(1-\cos 2\theta )+4\sin \theta )d\theta }$
${\displaystyle =4{\int }_{7\frac{\pi }{6}}^{3\frac{\pi }{2}}(3-2\cos 2\theta +4\sin \theta )d\theta }$
${\displaystyle =4[3\theta -\sin 2\theta +4{(-\cos \theta ]}_{7\frac{\pi }{6}}^{3\frac{\pi }{2}}}$
${\displaystyle =4[\frac{9\pi }{2}-\frac{7\pi }{2}-\frac{32}{\sqrt{3}}]}$
${\displaystyle A=4\pi -6\sqrt{3}}$